A review of $\eta$ efficiency factor for propellers

Question

A glider routinely creates 20x lift force than its propulsive force: the vertical component of gravity, which is sine angle of descent × weight. Is it's "efficiency" is greater than 1? Is it 20? Maybe more? It's also moving forward. Cdi + Cd0.

Is it possible that propellers, being airfoils, like wings, have an efficiency factor greater than 1? More like 4?

Let's test this with our Cessna 172R with an engine output torque of 350 footpounds spinning a 4 foot prop at 2700rpm. 180 horsepower.

Thrust = Drag

But wait, which drag? What comes first? Engine spins prop. Engine power output in footpounds = Prop drag in foot pounds. How much is this?

350 footpounds/0.75 × prop radius = 350/1.5 = 233 pounds of drag

The propeller is an airfoil, just like a wing, with a Lift/Drag ratio. Multiply by 4 and we have 932 pounds of thrust.

Discussion of "Power" as Force × distance/time for propellers

In the Cessna example 350 foot pounds yields 233 pounds of drag over a larger rotational distance than the engine crankshaft radius. (Imagine the force of the pistons!).

But it is the 233 pounds of drag which produces around 932 pounds of (horizontal lift) thrust.

How do airfoils multiply force?

force × distance/time

Simply put, the leading edge of the airfoil where force is applied to move air out of the way is a smaller distance than the chord of the wing. Lifting force is created over this chord above and below the wing, in the same amount of time.

Discussion of Thrust = $\eta\frac{P}{V}$

For steady state flight:

Thrust = Drag
$\eta$Power = Thrust × Velocity = Drag × Velocity = 1/2$\rho$V$^2$SCd × V

Thrust × Velocity = 1/2$\rho$V$^3$SCd

Thrust × Velocity = $\eta$Power

$\eta$Power/V = Thrust, Velocity factors out !

What is the efficiency factor?

is true efficiency force of pistons vs force of thrust?

Or even potential energy of fuel vs force of thrust per second, per hour etc.

Answer 1

Efficiency ($η$) Cannot Exceed 1

• Law of conservation of energy prohibits that.

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A glider routinely creates 20x lift force than its propulsive force: the vertical component of gravity, which is sine angle of descent × weight. Is it's "efficiency" is greater than 1? Is it 20? Maybe more?

• That's not its "efficiency" but the ratio of $\frac {Output\ Force}{Input\ Force}$ which is greater than 1. The same result can be obtained through reduction gears. Given enough leverage, you can obtain a million times more force (torque) than what you started with, but you can never obtain any more power than what you started with.

• This reduction gear system may produce a million times greater torque, but it will also inevitably have a million times less rotational speed, such that their product (power) remains the same.

• Similarly, a wing may produce much greater lift than drag, but the "lift power" of the wing is:

$\small{Lift\ power = Lift \cdot Velocity \cdot cosθ}$

• Since θ is $90°$ (lift is perpendicular to aircraft velocity), cosθ and thus lift power is zero. What is the propulsive efficiency of the wing? Zero - by definition.

Efficiency is not $\frac {Force\ Obtained}{Force\ Invested}$

efficiency is $\frac {Power\ Obtained}{Power\ Invested}$

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Propulsive Efficiency ($η$)

This is the efficiency of conversion of power invested (shaft power) into thrust power.

$\large{η = \frac {Thrust\ Power}{Shaft\ Power}}$

Propulsive efficiency can never be greater than 1, since that violates the law of conservation of energy. In practice, it will always be less than 1 since there will always be inefficiencies.

$\small{Thrust\ power = Speed \cdot Thrust}$

$\small{Shaft\ power = Rotation\ Speed \cdot Torque}$

Therefore:

$\large{η = \frac {Speed \cdot Thrust}{Rotation\ Speed \cdot Torque}}$

This is the correct expression for propulsive efficiency. Note that speed here is in TAS.

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Force and Power

In the question, it seems like it's being assumed that the L/D ratio of the propeller blades is its "efficiency". While it's true that both L/D ratio and $η$ are indicative of efficiency, both are fundamentally different:

L/D is a ratio of forces, η is a ratio of powers.

Let's understand $η$ graphically. For an engine operating at a constant power of 1 watt and an (unrealistic) propeller with $η = 1$, the thrust-speed graph¹ looks like this:

This is the thrust boundary that, for the given power, cannot be exceeded (let's call this the "limit curve"). As speed reduces to zero, thrust that a 1 watt engine can produce tends to infinity. The opposite is true as speed tends to c (speed of light). By definition, 1 watt of power at a speed of 1 $m/s$ yields a thrust of 1 Newton.

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Now let's compare this limit curve to the thrust-speed curve of a fixed pitch propeller driven by an engine at a constant power.

It can be seen that the limit curve is never trespassed, as prohibited by the law of conservation of energy.

It was earlier stated that $η = \frac{Thrust\ Power}{Shaft\ Power}$

$\large{η = \frac{T V}{P}}$

$\large{T = \frac {η P}{V}}$

If P and V are fixed, then $η$ only depends on T. Therefore, at a given speed:

$\large{η = \frac{T}{T'}}$

Where T' is the limit curve thrust for that speed.

Dividing T by T' is how we got the graph for $η$. The following features can be observed in this graph:

• At $V_0$, $η$ is (always) zero. That is because the maximum thrust that can be produced at $V_0$ (thrust at $η=1$) tends to infinity.

• $V_1$ is when $η$ is maximum - around 75% in this case. This is the cruise speed for which the propeller was designed - to provide maximum efficiency.

• At $V_2$, $η$ is zero again, because the blades have reached their $α_0$ (zero-lift AoA). The maximum speed that can be attained in level flight for the given engine power is always less than this speed².

• The blades stall at $V_3$. (there will also be a very high speed at which the blades will experience a negative-stall.)

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¹ _{This graph is not directly applicable to engines that carry their own propellant (rocket engines).}

² _{The exact maximum speed can be calculated using math from this answer.}

Answer 2

This question, which is remarkable, appears to largely relate to a misunderstanding of general theory. Consequently, we are caught on the horns of a dilemma: To what extent are we, as responsible persons, required to correct the misunderstandings formed of obvious misimpressions regarding the subject at hand? Understandably, the misinformed person will respond as tho being judged. Nevertheless, such is not our intent. We cannot possibly correct those misunderstandings unless the person holding them is willing to set them aside and objectively consider theory from a correct point of view.

The content of this answer is somewhat philosophical - not really engineering theory, not really science, perhaps unnecessary. But consider the following -

The supreme misfortune is when theory outstrips performance.
$ $ da Vinci

and

Nothing has a performance ratio of 1.

Both of these statements come from a) an observation of human nature and b) an observation of demonstrated fact. We cannot have a perpetual motion machine. A glider with a performance ratio of 1 would stay aloft indefinitely.

The performance ratio of a glider is given by its lift to drag ratio. Consequently a glider with a lift to drag ratio of 50 would have an efficiency of 98 percent. The impossibility is obvious if we consider adding just one more percent to the efficiency of a glider. The required lift to drag ratio is now 100, as of yet unachieved, and presently seen as unachievable in unassisted gliding flight.

Regarding propellers, my knowledge is rather limited. My general discoveries regarding propellers have shown me that there are some pretty good methods for calculating the thrust of a propeller, but that the most accessible and empirically based methods give only a reasonable estimate of propeller thrust and performance. What is the issue with propellers?

The propeller is usually the last thing considered in aircraft design. Why is that? When we get a design on paper, we essentially know all of the aspects of thrust and power required for flight. We get everything down to the powerplant and weight and balance, except for the propeller. Then we find we need to develop the required thrust by conversion of our powerplant output to real, useful work in propelling our aircraft. Now we need to select a propeller. We now have to determine if we build our own propeller from scratch, or purchase one already made, in a generic sense, from a reputable manufacturer. Aircraft design does not start with the propeller first, and then development of an airframe for fitting. It rather starts from the airframe first, and then deals with the propeller as a special problem. Elsewhere, Peter Kämpf and sophit have given excellent answers for consideration. Look at what they have said regarding propeller thrust calculations.

There is no desire here to lay out chapter and verse of flight theory in correcting a misunderstanding. But the following will be said: Powered flight comes down to three basic elements, namely weight, drag, and velocity. These elements are opposed in flight by lift, thrust, and power. Each of these elements has a manifested and interconnected relationship with the others, as follows -

static factor	aerodynamic requirement	the relationship
weight	lift	= (power/velocity)/(lift/drag)
drag	thrust	= weight/(lift/drag)
velocity	power	= velocity x thrust

If necessary, we can state these interrelationships in general algebraically defined terms. Skeptical?? Just write out the equations as equalities in the relations and see what happens as each element is changed in definition. Here, in basic terms, no equations yet, using the British engineering system, our expression

$$(lift^2/drag)*(velocity)=power$$ becomes $$lb * ft/s = power.$$ OMG!

So here are some thoughts. There are likely some who will read this answer and say it has no place on SE-Aviation. My leaning is in agreement with that sentiment. Nevertheless, the OP's questions, and answers, are valued. The hard part is learning thru developing the answers; teaching thru developing the questions.

Answer 3

The power consumed by drag in a gliding aircraft is simply its weight times its best rate of descent in feet per second. So a well-designed glider glides on less than 20 horsepower. This means to turn the glider into a plane, it would need an engine of more than 20 horsepower in order to climb.

Similarly, the power needed to climb is the weight of the plane times the desired vertical speed, plus the drag horsepower which can be found by solving for the glide equation above (and assuming the airspeed in your glide is equal to the airspeed in your ascent).

BTW If you tie one end of a tension gauge to the tail of a 172 and chain its other end down and firewall the engine, you will not register anywhere near 900 pounds of thrust...